Median String

You are given two strings ss and tt, both consisting of exactly kk lowercase Latin letters, ss is lexicographically less than tt.

Let's consider list of all strings consisting of exactly kk lowercase Latin letters, lexicographically not less than ss and not greater than tt (including ss and tt) in lexicographical order. For example, for k=2k=2, s=s="az" and t=t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].

Your task is to print the median (the middle element) of this list. For the example above this will be "bc".

It is guaranteed that there is an odd number of strings lexicographically not less than ss and not greater than tt.

Input

The first line of the input contains one integer kk (1≤k≤2⋅1051≤k≤2⋅105) — the length of strings.

The second line of the input contains one string ss consisting of exactly kk lowercase Latin letters.

The third line of the input contains one string tt consisting of exactly kk lowercase Latin letters.

It is guaranteed that ss is lexicographically less than tt.

It is guaranteed that there is an odd number of strings lexicographically not less than ss and not greater than tt.

Output

Print one string consisting exactly of kk lowercase Latin letters — the median (the middle element) of list of strings of length kk lexicographically not less than ss and not greater than tt.

Examples

Input

2
az
bf

Output

bc

Input

5
afogk
asdji

Output

alvuw

Input

6
nijfvj
tvqhwp

Output

qoztvz

题目大致题意就是给你两个字符串数组，让你用计算他们的中间值，即转化为26进制后的中间值，
我的想法就是把每一位进行两个数组的加法，通过每一位的进位来计算，对每一个数值mod26，如果大于26，就向前进一位
最后计算的时候如果是偶数就是直接取中间值，如果奇数就想下一位；
接下来就是代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define ll long long 
const int maxn = 200100;
const int inf = 0x3f3f3f3f;
char a[maxn], b[maxn];
int c[maxn], d[maxn],e[maxn];
int main()
{
    int k;
    cin >> k;
    cin >> a;
    cin >> b;
    for (int i = k-1;i >=0 ;i--)
    {
        c[i] = a[i] - 'a';
        d[i] = b[i] - 'a' ;
    }
    memset(e, 0, sizeof(e));
    for (int i = k - 1;i >= 0;i--)
    {
        e[i] += c[i] + d[i];
        if (e[i] >= 26&&i!=0)
        {
            e[i] = e[i] % 26;
            e[i - 1] += 1;
            
        }
    }
    for (int  i = 0; i < k; i++)
    {
        if (e[i] % 2 == 0)
        {
            printf("%c", e[i] / 2 + 'a');
        }
        else
        {
            printf("%c", e[i] / 2 + 'a');
            e[i + 1] += 26;
        }
    }
}